close
標題:
binomial theorem
發問:
最佳解答:
( 1 + ax + 3x^2 + 5x^3 )( 1 - 1/x )^5 =( 1 + ax + 3x^2 + 5x^3 )[ 1 - 5C1 (1/x) + 5C2 (1/x)^2 - 5C3(1/x)^3 + ...] =( 1 + ax + 3x^2 + 5x^3 )(1 - 5/x + 10/x^2 + ... ] coefficient of x =a(1) + 3(-5) + 5(10) =a-15+50 =a+35 a+35=33 a= -2
其他解答:
binomial theorem
發問:
此文章來自奇摩知識+如有不便請留言告知
In the expansion of ( 1 + ax + 3x^2 + 5x^3 )( 1 - 1/x )^5 ,the coefficient of x is 33.Find the value of a. 更新: 答得詳細d ..得唔得........ 好似少左好多步驟最佳解答:
( 1 + ax + 3x^2 + 5x^3 )( 1 - 1/x )^5 =( 1 + ax + 3x^2 + 5x^3 )[ 1 - 5C1 (1/x) + 5C2 (1/x)^2 - 5C3(1/x)^3 + ...] =( 1 + ax + 3x^2 + 5x^3 )(1 - 5/x + 10/x^2 + ... ] coefficient of x =a(1) + 3(-5) + 5(10) =a-15+50 =a+35 a+35=33 a= -2
其他解答:
文章標籤
全站熱搜
留言列表
